consider the situastion of the previous problem. Suppose the block of mass `m_1` is pulled by a constant force `F_1` and the other block is pulled by a constant force `F_2`. Find the maximum elongation that the spring will suffer.

Answer

A, B

Explanation

Acceleration of mass `m_1=(F_1-|F_2)/(m_1+m-2)`
Similarly acceleration of mass ltbr. `m_2=(F_2-F_1)/(m_1+m_2)`
due to `F_1 and F_2` block of mass `m_1 and m_2` will expereince different acceleration and experience an ineertia force
`:.` Net forc eon `m_1=F_1-m_1a`
`=F_1-m_1xx((F_1-F_2))/(m-1+m-2)`
`(m_1F_1+m_2F_1-m_1F_1+_2m-1)/(m-1+m_2)`
`(m_2F_1+m-1F_2)/(m_1+m_2)`
Similarly net force on `m_2`
`=F_2-m_2xx((F_2-F_1))/(m_1+m_2)`
`=(m_1F_2+m_2F_2-m_2F_2+m_2F_1)/(m_1+m_2)`
`=(m_1.F_2+m_2.F_2)/(m_1+m_2)`
`:.If m_1` is dispaced by a distance `x_1 and x_2 by m_2` the maximum extension of the spring is `x_1+x_2`.
`:.` work done by the bocks =energy stored in the spring.
`rarr (m_2.F_1+m_1F_2)/(m_1+m_2)x(x_1)+(m_2F_1+m_1F_2)/(m_1+m_2)xx_2`
`=(1/2)K(x_1+x_2)^2`
`rarr x_1+x_2=2/k (m_2F_1+m_1F_2)/(m_1+m_2)`